Question 446: Electrostats & Capacitance

Question

The capacity of air filled parallel plate capacitor is $\mathrm{C}_0$. One-half of the space between the plates is filled with a dielectric constant ' K ' as shown in figure. The new capacity becomes $\mathrm{C}_{\mathrm{n}}$. The ratio $\mathrm{C}_{\mathrm{n}}$ to $\mathrm{C}_0$ is

Accepted Answer

Original capacitor (air-filled): $\mathrm{C}_0$
After adding dielectric $(\mathrm{K})$ to half the plate area new capacitance $=\mathrm{C}$
$$ C_0=\frac{\varepsilon_0 A}{d} $$
The plate is split in to two halves in parallel:
One half with air: A/2, dielectric constant = 1
One half with dielectric K : $\mathrm{A} / 2$, dielectric constant $=K$
$$ \begin{aligned} &\begin{array}{ll} 	herefore & C_{	ext {air }}=\frac{E_0(A / 2)}{d}=\frac{1}{2} \cdot \frac{\varepsilon_0 A}{d}=\frac{C_0}{2} \ 	herefore & C_{	ext {dielectric }}=\frac{K \varepsilon_0(A / 2)}{d}=\frac{K}{2} \cdot \frac{\varepsilon_0 A}{d}=\frac{K C_0}{2} \end{array}\ &	ext { Total Capacitance (Parallel Combination) }\ &\begin{aligned} & C=C_{air}+C_{dielectric}=\frac{C_0}{2}+\frac{K C_0}{2}=\frac{C_0(1+K)}{2} \ 	herefore & \frac{C_0}{C_0}=\left(\frac{K+1}{2}ight) \end{aligned} \end{aligned} $$

Answer

$\left(\frac{\mathrm{K}+1}{2}\right)$

Answer

$\left(\frac{\mathrm{K}+1}{3}\right)$

Answer

$\left(\frac{\mathrm{K}+1}{4}\right)$

Answer

$4(\mathrm{~K}+1)$

Electrostats & Capacitance

The capacity of air filled parallel plate capacitor is \mathrm{C}_0. One-half of the space between the plates is filled with a dielectric constant ' K ' as shown in figure. The new capacity becomes \mathrm{C}_{\mathrm{n}}. The ratio \mathrm{C}_{\mathrm{n}} to \mathrm{C}_0 is

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